Are there "simple" functions whose asymptotic behaviour is impossible to know?

Question

My question is

Does exist a simple function $f \colon A \subseteq \mathbb R \to \mathbb R$ such that it is impossible to know $\lim_{x \to \infty} f(x)$ ?

Of course there are a lot of function whose behaviour is not known; but what happens if we use only elementary functions (the rigourous definition of elementary function involves differential algebra, so let's just imagine a elementary function as a composition of exponentials, logarithms, rational functions and trigonometric functions)?. More clearly the problem is:

Is it possible to create some function (or sequence) "easy to define" for which is not possible to evaluate his asymptotic behaviour?

Edit: with "not possible to evaluate" I mean that the problem to evaluate $$\lim_{x \to +\infty} f(x)$$ is not decidible.

Answer 1

Whatever you mean, the following function satisfies your request: $$f(x)= \begin{cases} x & \mbox{ if the Riemann Hypothesis is true}\\ -x & \mbox{ otherwise}\end{cases}$$

Knowing the limit of this function is not impossible, but it's still hard, since it's equivalent on proving the Riemann Hypothesis.

If you want something actually impossible, substitute the Riemann Hypothesis with something impossible to prove, for example the Continuum Hypothesis.

Answer 2

Consider a sequence defined as follows (also see here):

$a_1=1$

$a_{n+1}\in\{a_n,a_n+1\}$

$a_n=$ number if occurences of $n$ in the sequence.

The recurrence defined above implies

$a_n+1=a_{n+a_{a_n}}$ Eq. 1

Suppose that we assume a leading behavior with the form

$a_n\approx \alpha n^\beta$ Eq. 2

Plugging Eq. 2 into Eq. 1 and applying the binomial power expansion to the right side gives what turns out to be the correct leading behavior:

$\alpha=\phi^{1/\phi^2}$

$\beta=1/\phi$

$\phi=(1+\sqrt5)/2$

This, put into Eq. 2, is indeed the correct leading behavior. But ... the higher order correction terms coming from the binomial expansion are $O(1)$, hence cannot rise above the "noise" generated by the unit jumps in this sequence of whole numbers. Therefore, the leading behavior cannot be expanded into a power series or refined in any other way using analytic functions.

Answer 3

This is an wild attempt.

More than as an answer, it might be thought as a way to get closer to a constructive answer.

Let $P(n)$ be a statement which depends on $n$ with two properties:

• for every $n$, it is possible to determine whether $P(n)$ is true or false
• the statement "there exists $n_0$ for which $P(n)$ is true for every $n > n_0$" is undecidable

Then define $f : \mathbb{N} \to \mathbb{R}$ by $$ f(n) =\sum_{\substack{m < n , \\ P(m) \text{ is true}}} \frac{1}{m} $$

Then (depending on $P$) I would expect that convergence or divergence of $f(n)$ is undecidable (and tweaking with some signs one can make it into a problem about existence of the limit).

However, this $f$ does not really give an answer to the question without an explicit instance of $P$, because there is an obvious problem if $P(n)$ is true for enough $n$ to have the series diverge, but it is not necessarily true for all $n$.

Hope this helps a bit.